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An elevator in a mine ascends at a rate of 3 metre / minute anddescends at a rate of 5 metre/minute. The distance above the ground level represents a positve integerand below the ground level represents a negative integer. If eleveator is at a height of 30 metresabove the ground. Find the position of elevator(a) When it descends for 30 minutes(b) When it ascends for 10 minnutes(C) Find the position of elevator when it starts from ground level and descends for 40 minutes andafter that ascends for 20 minutes. |
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Answer» Ncert solutions Grade 7 Mathematics Science Chapters in NCERT Solutions - Mathematics, Class 7 Exercises in Integers Question 7 Q7) An elevator descends into a mine shaft at the rate of 6 m/MIN. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m. Solution Transcript Solution 7: Starting position of mine shaft is = 10m above ground but, it moves in opposite direction so it travels the DISTANCE (-350 m ) below the ground. Total distance covered by mine shaft = 10m - (-350m) = 10 + 350 = 360m Now, taken to cover a distance of 6m by it = 1 minute Time taken to cover a distance of 1m by it = \frac{1}{6\ }\min 6 1
min Therefore, tie taken to a cover a distance of 360m = \frac{1}{6}\TIMES\ 360\ =\ 60\ \min\ \ 6 1
× 360 = 60 min 60 minutes = 1 hour thus, in one hour the mine shraft reaches = 350 below the ground. BY TEJAS KULKARNI PLEASE FOLLOW AND MARK AS BRAINLIST |
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